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In the oxidation of ethane: 2 C2H6 + 7 02 → 4 CO2 + 6 H2O how many
liters of O2 are required to react with 90 grams of ethane?

Sagot :

Eduard22sly

Answer:

10.5 L of O₂.

Explanation:

We'll begin by calculating the number of mole in 90 g of C₂H₆. This can be obtained as follow:

Molar mass of C₂H₆ = (12×2) + (6×1)

= 24 + 6

= 30 g/mol

Mass of C₂H₆ = 90 g

Mole of C₂H₆ =?

Mole = mass / molar mass

Mole of C₂H₆ = 90 / 30

Mole of C₂H₆ = 3 moles.

Finally, we shall determine the volume of O₂ required. This is illustrated:

2C₂H₆ + 7O₂ —> 4CO₂ + 6H₂O

From the balanced equation above, we can say that:

2 L of C₂H₆ required 7 L of O₂.

Therefore, 3 L of C₂H₆ will require

= (3 × 7)/ 2 = 10.5 L of O₂.

Thus, 10.5 L of O₂ is required for the reaction.

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